3.276 \(\int \cot (c+d x) (a+b \sec (c+d x))^2 \, dx\)

Optimal. Leaf size=61 \[ \frac{a^2 \log (\cos (c+d x))}{d}+\frac{(a+b)^2 \log (1-\sec (c+d x))}{2 d}+\frac{(a-b)^2 \log (\sec (c+d x)+1)}{2 d} \]

[Out]

(a^2*Log[Cos[c + d*x]])/d + ((a + b)^2*Log[1 - Sec[c + d*x]])/(2*d) + ((a - b)^2*Log[1 + Sec[c + d*x]])/(2*d)

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Rubi [A]  time = 0.0996327, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {3885, 1802} \[ \frac{a^2 \log (\cos (c+d x))}{d}+\frac{(a+b)^2 \log (1-\sec (c+d x))}{2 d}+\frac{(a-b)^2 \log (\sec (c+d x)+1)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]*(a + b*Sec[c + d*x])^2,x]

[Out]

(a^2*Log[Cos[c + d*x]])/d + ((a + b)^2*Log[1 - Sec[c + d*x]])/(2*d) + ((a - b)^2*Log[1 + Sec[c + d*x]])/(2*d)

Rule 3885

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[((b^2 - x^2)^((m - 1)/2)*(a + x)^n)/x, x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin{align*} \int \cot (c+d x) (a+b \sec (c+d x))^2 \, dx &=-\frac{b^2 \operatorname{Subst}\left (\int \frac{(a+x)^2}{x \left (b^2-x^2\right )} \, dx,x,b \sec (c+d x)\right )}{d}\\ &=-\frac{b^2 \operatorname{Subst}\left (\int \left (\frac{(a+b)^2}{2 b^2 (b-x)}+\frac{a^2}{b^2 x}-\frac{(a-b)^2}{2 b^2 (b+x)}\right ) \, dx,x,b \sec (c+d x)\right )}{d}\\ &=\frac{a^2 \log (\cos (c+d x))}{d}+\frac{(a+b)^2 \log (1-\sec (c+d x))}{2 d}+\frac{(a-b)^2 \log (1+\sec (c+d x))}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.10629, size = 53, normalized size = 0.87 \[ \frac{(a+b)^2 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+(a-b)^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )-b^2 \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]*(a + b*Sec[c + d*x])^2,x]

[Out]

((a - b)^2*Log[Cos[(c + d*x)/2]] - b^2*Log[Cos[c + d*x]] + (a + b)^2*Log[Sin[(c + d*x)/2]])/d

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Maple [A]  time = 0.039, size = 53, normalized size = 0.9 \begin{align*}{\frac{{b}^{2}\ln \left ( \tan \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{ab\ln \left ( \csc \left ( dx+c \right ) -\cot \left ( dx+c \right ) \right ) }{d}}+{\frac{{a}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)*(a+b*sec(d*x+c))^2,x)

[Out]

1/d*b^2*ln(tan(d*x+c))+2/d*a*b*ln(csc(d*x+c)-cot(d*x+c))+1/d*a^2*ln(sin(d*x+c))

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Maxima [A]  time = 0.984351, size = 84, normalized size = 1.38 \begin{align*} -\frac{2 \, b^{2} \log \left (\cos \left (d x + c\right )\right ) -{\left (a^{2} - 2 \, a b + b^{2}\right )} \log \left (\cos \left (d x + c\right ) + 1\right ) -{\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (\cos \left (d x + c\right ) - 1\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

-1/2*(2*b^2*log(cos(d*x + c)) - (a^2 - 2*a*b + b^2)*log(cos(d*x + c) + 1) - (a^2 + 2*a*b + b^2)*log(cos(d*x +
c) - 1))/d

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Fricas [A]  time = 0.873844, size = 184, normalized size = 3.02 \begin{align*} -\frac{2 \, b^{2} \log \left (-\cos \left (d x + c\right )\right ) -{\left (a^{2} - 2 \, a b + b^{2}\right )} \log \left (\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right ) -{\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (-\frac{1}{2} \, \cos \left (d x + c\right ) + \frac{1}{2}\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/2*(2*b^2*log(-cos(d*x + c)) - (a^2 - 2*a*b + b^2)*log(1/2*cos(d*x + c) + 1/2) - (a^2 + 2*a*b + b^2)*log(-1/
2*cos(d*x + c) + 1/2))/d

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \sec{\left (c + d x \right )}\right )^{2} \cot{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*(a+b*sec(d*x+c))**2,x)

[Out]

Integral((a + b*sec(c + d*x))**2*cot(c + d*x), x)

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Giac [A]  time = 1.38499, size = 136, normalized size = 2.23 \begin{align*} -\frac{2 \, a^{2} \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) + 2 \, b^{2} \log \left ({\left | -\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) -{\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (\frac{{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)*(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

-1/2*(2*a^2*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) + 2*b^2*log(abs(-(cos(d*x + c) - 1)/(cos(d*x
+ c) + 1) - 1)) - (a^2 + 2*a*b + b^2)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)))/d